Show that $\min(X,Y)$ has geometric distribution

$X,Y$ - random variables,
$X-Geom(p)$
$Y-Geom(r)$
$$P(X=k) = (1-p)^{k-1}p$$ $$P(Y=k) = (1-r)^{k-1}r$$ $X,Y$ are independent. Show that $Z=\min(X,Y)-Geom(t)$

My attempt: $P(\min(X,Y))=P(X=k\wedge Y\ge k) + P(Y=k\wedge X\ge k) - P(Y=k\wedge X=k) = P(X=k)\cdot P(Y\ge k) + P(Y=k)\cdot P(X\ge k) - P(X=k)\cdot P(Y=k)$ And I don't idea what can I do now to finish it.

Solution 1:

With the minimum, a bit of cleverness is necessary: $$\begin{align} \mathbb{P}\left(Z \leq z\right) = \mathbb{P}\left(\min(X, Y) \leq z\right) &= 1 - \mathbb{P}\left(\min(X, Y) > z\right) \\ &= 1 - \mathbb{P}\left(\text{both }X \text{ and }Y > z\right) \\ &= 1 - \mathbb{P}(X > z)\mathbb{P}(Y>z)\text{.} \end{align}$$ Note the above is the distribution function of $Z$. Now $$\mathbb{P}(X > z) = \sum\limits_{k = z+1}^{\infty}(1-p)^{k-1}p = p[(1-p)^{z}+(1-p)^{z+1}+\cdots] = p(1-p)^{z}\left[\dfrac{1}{1-(1-p)}\right] = (1-p)^{z}\text{, } \quad |1-p|<1\text{.}$$ You can find a similar result for $\mathbb{P}(Y > z)$.

Use the resulting equation of $\mathbb{P}(Z \leq z)$ to derive an explicit formula for $\mathbb{P}\left(Z = z\right)$ by using that $$\mathbb{P}\left(Z = z\right) = \mathbb{P}\left(Z \leq z\right) - \mathbb{P}\left(Z \leq z-1\right)\text{.}$$